Arturo O. answered • 10/31/17
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(i)
From kinematics,
v(t) = v0 + at = v0 - gt
v(t) = 0 at maximum height ⇒
0 = v0 - gt
v0 = gt = (9.8 m/s2)(4 s) = 39.2 m/s
(ii)
The horizontal distance traveled is zero.
h(t) = -(1/2)gt2 + v0t + h0 = -4.9t2 + 39.2t + 2
h(4) = distance traveled (in meters) on the way up
h(4) + 2 = distance traveled (in meters) on the way down
h(4) + [h(4) + 2] = total vertical distance traveled. Can you finish from here?
Arturo O.
Kela,
h(4) = [-4.9(42) + 39.2(4) + 2] m = ? m
The result is the distance traveled on the way up. But since it launched from a height of 2 m, the distance traveled on the way to the ground is h(4) + 2 m. Just add the distance going up to the distance coming down and that is the total distance traveled in the vertical direction (the horizontal distance is zero since it was launched straight up). You have all of the equations. It is now a matter of just plugging in the numbers, which you know how to do.
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11/01/17
Kela W.
11/01/17